- #1
TerryW
Gold Member
- 229
- 21
- Homework Statement
- Derive Equation (15) from Equation (12)
- Relevant Equations
- See attachment
I haven't posted for a while and I am still (!) working through some of the things I didn't quite get in MTW Chapter 21.
Here is my latest puzzle.
I want to work out how to get from Equation (12) in the attachment, to Equation (15).
I've tried the "add and subtract" ##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_0\delta t\}_{,i}##
This gives me ##+\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta t\}A_{0,i}## and -##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta t\}A_{i,0}##
Plus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t## and minus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t##
All this does is allow me to replace ##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{i,0}\}## with ##-\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{0,i}\}## which I could have done anyway by index manipulation,
I can then add the two versions of (12) to give a new equation which is $$2\delta S = \int \big[ 2\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta A_{i}+\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{i,0}-\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{0,i}\}\delta t -2\mathfrak L\}\big]d^3x$$
What this means is that my result for ##\frac {\delta S}{\delta \Omega}## contains the term ##2F^{i0}(A_{i,0} - A_{0,i})## instead of ##4F^{i0}(A_{i,0} - A_{0,i})##
I then had a look at the Plus and Minus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t## terms which I had discarded earlier as they cancel, to see if I could find some extra terms, but I couldn't find anything to fix the problem.
Can anyone point out what I am missing?
RegardsTerryW
Here is my latest puzzle.
I want to work out how to get from Equation (12) in the attachment, to Equation (15).
I've tried the "add and subtract" ##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_0\delta t\}_{,i}##
This gives me ##+\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta t\}A_{0,i}## and -##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta t\}A_{i,0}##
Plus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t## and minus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t##
All this does is allow me to replace ##\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{i,0}\}## with ##-\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{0,i}\}## which I could have done anyway by index manipulation,
I can then add the two versions of (12) to give a new equation which is $$2\delta S = \int \big[ 2\frac {(-g)^{\frac12}F^{i0}}{4\pi}\delta A_{i}+\{\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{i,0}-\frac {(-g)^{\frac12}F^{i0}}{4\pi}A_{0,i}\}\delta t -2\mathfrak L\}\big]d^3x$$
What this means is that my result for ##\frac {\delta S}{\delta \Omega}## contains the term ##2F^{i0}(A_{i,0} - A_{0,i})## instead of ##4F^{i0}(A_{i,0} - A_{0,i})##
I then had a look at the Plus and Minus ## \{\frac {(-g)^{\frac12}F^{i0}}{4\pi}\}_{,i}A_0\delta t## terms which I had discarded earlier as they cancel, to see if I could find some extra terms, but I couldn't find anything to fix the problem.
Can anyone point out what I am missing?
RegardsTerryW
